lecture 9 ee

14

Alternating voltages and currents

At the end of this chapter you should be able to:

ž appreciate why a.c. is used in preference to d.c.

ž describe the principle of operation of an a.c. generator

ž distinguish between unidirectional and alternating waveforms

ž define cycle, period or periodic time T and frequency f of a waveform

ž perform calculations involving T D 1/f

ž define instantaneous, peak, mean and r.m.s. values, and form and peak factors for asine wave

ž calculate mean and r.m.s. values and form and peak factors for given waveforms

ž understand and perform calculations on the general sinusoidal equationv D Vm sin⊲ωt š �⊳

ž understand lagging and leading angles

ž combine two sinusoidal waveforms (a) by plotting graphically, (b) by drawingphasors to scale and (c) by calculation

14.1 Introduction

Electricity is produced by generators at power sta-tions and then distributed by a vast network oftransmission lines (called the National Grid system)to industry and for domestic use. It is easier andcheaper to generate alternating current (a.c.) thandirect current (d.c.) and a.c. is more convenientlydistributed than d.c. since its voltage can be readilyaltered using transformers. Whenever d.c. is neededin preference to a.c., devices called rectifiers areused for conversion (see Section 14.7).

14.2 The a.c. generator

Let a single turn coil be free to rotate at constantangular velocity symmetrically between the polesof a magnet system as shown in Fig. 14.1

Figure 14.1

An e.m.f. is generated in the coil (from Faraday’slaws) which varies in magnitude and reverses itsdirection at regular intervals. The reason for this isshown in Fig. 14.2 In positions (a), (e) and (i) theconductors of the loop are effectively moving alongthe magnetic field, no flux is cut and hence no e.m.f.is induced. In position (c) maximum flux is cut and

TLFeBOOK

184 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

Figure 14.2

hence maximum e.m.f. is induced. In position (g),maximum flux is cut and hence maximum e.m.f. isagain induced. However, using Fleming’s right-handrule, the induced e.m.f. is in the opposite directionto that in position (c) and is thus shown as E. Inpositions (b), (d), (f) and (h) some flux is cut andhence some e.m.f. is induced. If all such positionsof the coil are considered, in one revolution of thecoil, one cycle of alternating e.m.f. is produced asshown. This is the principle of operation of the a.c.generator (i.e. the alternator).

14.3 Waveforms

If values of quantities which vary with time t areplotted to a base of time, the resulting graph is calleda waveform. Some typical waveforms are shown inFig. 14.3. Waveforms (a) and (b) are unidirectionalwaveforms, for, although they vary considerablywith time, they flow in one direction only (i.e. theydo not cross the time axis and become negative).Waveforms (c) to (g) are called alternating wave-forms since their quantities are continually changingin direction (i.e. alternately positive and negative).

A waveform of the type shown in Fig. 14.3(g) iscalled a sine wave. It is the shape of the waveformof e.m.f. produced by an alternator and thus themains electricity supply is of ‘sinusoidal’ form.

One complete series of values is called a cycle(i.e. from O to P in Fig. 14.3(g)).

The time taken for an alternating quantity tocomplete one cycle is called the period or theperiodic time, T, of the waveform.

The number of cycles completed in one secondis called the frequency, f, of the supply and is

Figure 14.3

measured in hertz, Hz. The standard frequency ofthe electricity supply in Great Britain is 50 Hz

T =

1

for f =

1

T

Problem 1. Determine the periodic time forfrequencies of (a) 50 Hz and (b) 20 kHz.

(a) Periodic time T D1

fD

1

50D 0.02 s or 20 ms

(b) Periodic time T D1

fD

1

20 000

D 0.00005 s or 50ms

Problem 2. Determine the frequencies forperiodic times of (a) 4 ms (b) 4 µs.

(a) Frequency f D1

TD

1

4ð 10 3

D

1000

4D 250 Hz

(b) Frequency f D1

TD

1

4ð 10 6D

1 000 000

4

D 250 000 Hz

or 250 kHz or 0.25 MHz

Problem 3. An alternating currentcompletes 5 cycles in 8 ms. What is itsfrequency?

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ALTERNATING VOLTAGES AND CURRENTS 185

Time for 1 cycle D ⊲8/5⊳ ms D 1.6 ms D periodictime T.

Frequency f D1

TD

1

1.6ð 10 3D

1000

1.6

D10 000

16D 625 Hz

Now try the following exercise

Exercise 73 Further problems onfrequency and periodic time

1 Determine the periodic time for the followingfrequencies:

(a) 2.5 Hz (b) 100 Hz (c) 40 kHz[(a) 0.4 s (b) 10 ms (c) 25 µs]

2 Calculate the frequency for the following peri-odic times:

(a) 5 ms (b) 50 µs (c) 0.2 s[(a) 200 Hz (b) 20 kHz (c) 5 Hz]

3 An alternating current completes 4 cycles in5 ms. What is its frequency? [800 Hz]

14.4 A.c. values

Instantaneous values are the values of the alternat-ing quantities at any instant of time. They are repre-sented by small letters, i, v, e, etc., (see Fig. 14.3(f)and (g)).

The largest value reached in a half cycle is calledthe peak value or the maximum value or thecrest value or the amplitude of the waveform.Such values are represented by Vm, Im, Em, etc.(see Fig. 14.3(f) and (g)). A peak-to-peak value ofe.m.f. is shown in Fig. 14.3(g) and is the differencebetween the maximum and minimum values in acycle.

The average or mean value of a symmetricalalternating quantity, (such as a sine wave), is theaverage value measured over a half cycle, (sinceover a complete cycle the average value is zero).

Average ormean value

}

Darea under the curve

length of base

The area under the curve is found by approxi-mate methods such as the trapezoidal rule, the mid-ordinate rule or Simpson’s rule. Average values arerepresented by VAV, IAV, EAV, etc.

For a sine wave:

average value = 0.637 × maximum value

.i.e. 2=p × maximum value/

The effective value of an alternating current isthat current which will produce the same heatingeffect as an equivalent direct current. The effectivevalue is called the root mean square (r.m.s.) valueand whenever an alternating quantity is given, itis assumed to be the rms value. For example, thedomestic mains supply in Great Britain is 240 V andis assumed to mean ‘240 V rms’. The symbols usedfor r.m.s. values are I, V, E, etc. For a non-sinusoidalwaveform as shown in Fig. 14.4 the r.m.s. value isgiven by:

I D

√

i21 C i2

2 C . . .C i2n

n

where n is the number of intervals used.

Figure 14.4

For a sine wave:

rms value = 0.707 × maximum value

.i.e. 1=p

2 × maximum value/

Form factor =

r.m.s. value

average value

For a sine wave, form factor D 1.11

TLFeBOOK


Peak factor =

maximum value

r.m.s. value

For a sine wave, peak factor D 1.41.The values of form and peak factors give an

indication of the shape of waveforms.

Problem 4. For the periodic waveformsshown in Fig. 14.5 determine for each:(i) frequency (ii) average value over half acycle (iii) r.m.s. value (iv) form factor and(v) peak factor.

Figure 14.5

(a) Triangular waveform (Fig. 14.5(a)).

(i) Time for 1 complete cycle D 20 ms D

periodic time, T. Hence

frequency f D1

TD

1

20ð 10 3

D1000

20D 50 Hz

(ii) Area under the triangular waveform for a

half-cycle D 12ð baseð height

D12ð ⊲10ð 10 3⊳ð 200 D 1 volt second

Average valueof waveform

}

Darea under curve

length of base

D1 volt second

10ð 10 3second

D1000

10D 100 V

(iii) In Fig. 14.5(a), the first 1/4 cycle is dividedinto 4 intervals. Thus

rms value D

√

v21 C v

22 C v

23 C v

24

4

D

√

252C752

C1252C1752

4

D 114.6 V

(Note that the greater the number of inter-vals chosen, the greater the accuracy of theresult. For example, if twice the number ofordinates as that chosen above are used, ther.m.s. value is found to be 115.6 V)

(iv) Form factor Dr.m.s. value

average value

D114.6

100D 1.15

(v) Peak factor Dmaximum value

r.m.s. value

D200

114.6D 1.75

(b) Rectangular waveform (Fig. 14.5(b)).

(i) Time for 1 complete cycle D 16 ms D

periodic time, T. Hence

frequency, f D1

TD

1

16ð 10 3D

1000

16

D 62.5 Hz

(ii)Average value over

half a cycle

}

Darea under curve

length of base

D10ð ⊲8ð 10 3⊳

8ð 10 3

D 10 A

(iii) The r.m.s. value D

√

i21 C i2

2 C i23 C i2

4

4D 10 A, however many intervals are chosen,since the waveform is rectangular.

TLFeBOOK


(iv) Form factor Dr.m.s. value

average valueD

10

10D 1

(v) Peak factor Dmaximum value

r.m.s. valueD

10

10D 1

Problem 5. The following table gives thecorresponding values of current and time fora half cycle of alternating current.

time t (ms) 0 0.5 1.0 1.5 2.0current i (A) 0 7 14 23 40

time t (ms) 2.5 3.0 3.5 4.0 4.5 5.0current i (A) 56 68 76 60 5 0

Assuming the negative half cycle is identicalin shape to the positive half cycle, plot thewaveform and find (a) the frequency of thesupply, (b) the instantaneous values ofcurrent after 1.25 ms and 3.8 ms, (c) the peakor maximum value, (d) the mean or averagevalue, and (e) the r.m.s. value of thewaveform.

The half cycle of alternating current is shown plottedin Fig. 14.6

(a) Time for a half cycle D 5 ms; hence the time for1 cycle, i.e. the periodic time,T D 10 ms or 0.01 s

Frequency, f D1

TD

1

0.01D 100 Hz

(b) Instantaneous value of current after 1.25 ms is19 A, from Fig. 14.6. Instantaneous value ofcurrent after 3.8 ms is 70 A, from Fig. 14.6

(c) Peak or maximum value D 76 A

(d) Mean or average value Darea under curve

length of base

Using the mid-ordinate rule with 10 intervals,each of width 0.5 ms gives:

area undercurve

}

D ⊲0.5ð 10 3⊳[3C 10C 19C 30

C 49C 63C 73C 72C 30C 2]

(see Fig. 14.6)

D ⊲0.5ð 10 3⊳⊲351⊳

Figure 14.6

Hence mean oraverage value

}

D⊲0.5ð 10 3⊳⊲351⊳

5ð 10 3

D 35.1 A

(e) R.m.s value D

√

√

√

√

√

√

32C 102

C 192C 302

C 492C632

C732C722

C 302C 22

10

D

√

19157

10D 43.8 A

Problem 6. Calculate the r.m.s. value of asinusoidal current of maximum value 20 A.

For a sine wave,

r.m.s. value D 0.707ðmaximum value

D 0.707ð 20 D 14.14 A

Problem 7. Determine the peak and meanvalues for a 240 V mains supply.

TLFeBOOK


For a sine wave, r.m.s. value of voltageV D 0.707 ð Vm.A 240 V mains supply means that 240 V is the r.m.s.value, hence

Vm D

V

0.707D

240

0.707D 339.5 V

D peak value

Mean value

VAV D 0.637 Vm D 0.637 ð 339.5 D 216.3 V

Problem 8. A supply voltage has a meanvalue of 150 V. Determine its maximumvalue and its r.m.s. value.

For a sine wave, mean value D 0.637 ð maximumvalue. Hence

maximum value D

mean value

0.637D

150

0.637

D 235.5 V

R.m.s. value D 0.707 ð maximum valueD 0.707 ð 235.5 D 166.5 V


Exercise 74 Further problems on a.c.values of waveforms

1 An alternating current varies with time overhalf a cycle as follows:

Current (A) 0 0.7 2.0 4.2 8.4time (ms) 0 1 2 3 4

Current (A) 8.2 2.5 1.0 0.4 0.2 0time (ms) 5 6 7 8 9 10

The negative half cycle is similar. Plot thecurve and determine:(a) the frequency (b) the instantaneous valuesat 3.4 ms and 5.8 ms (c) its mean value and(d) its r.m.s. value[(a) 50 Hz (b) 5.5 A, 3.4 A (c) 2.8 A (d) 4.0 A]

2 For the waveforms shown in Fig. 14.7 deter-mine for each (i) the frequency (ii) the averagevalue over half a cycle (iii) the r.m.s. value(iv) the form factor (v) the peak factor.

[(a) (i) 100 Hz (ii) 2.50 A (iii) 2.88 A

(iv) 1.15 (v) 1.74

(b) (i) 250 Hz (ii) 20 V (iii) 20 V

(iv) 1.0 (v) 1.0

(c) (i) 125 Hz (ii) 18 A (iii) 19.56 A

(iv) 1.09 (v) 1.23

(d) (i) 250 Hz (ii) 25 V (iii) 50 V

(iv) 2.0 (v) 2.0]

Figure 14.7

3 An alternating voltage is triangular in shape,rising at a constant rate to a maximum of300 V in 8 ms and then falling to zero at aconstant rate in 4 ms. The negative half cycleis identical in shape to the positive half cycle.Calculate (a) the mean voltage over half acycle, and (b) the r.m.s. voltage

[(a) 150 V (b) 170 V]

4 An alternating e.m.f. varies with time over halfa cycle as follows:

E.m.f. (V) 0 45 80 155time (ms) 0 1.5 3.0 4.5

E.m.f. (V) 215 320 210 95time (ms) 6.0 7.5 9.0 10.5

E.m.f. (V) 0time (ms) 12.0

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The negative half cycle is identical in shapeto the positive half cycle. Plot the waveformand determine (a) the periodic time and fre-quency (b) the instantaneous value of voltageat 3.75 ms (c) the times when the voltage is125 V (d) the mean value, and (e) the r.m.s.value

[(a) 24 ms, 41.67 Hz (b) 115 V(c) 4 ms and 10.1 ms (d) 142 V

(e) 171 V]

5 Calculate the r.m.s. value of a sinusoidal curveof maximum value 300 V [212.1 V]

6 Find the peak and mean values for a 200 Vmains supply [282.9 V, 180.2 V]

7 Plot a sine wave of peak value 10.0 A. Showthat the average value of the waveform is6.37 A over half a cycle, and that the r.m.s.value is 7.07 A

8 A sinusoidal voltage has a maximum value of120 V. Calculate its r.m.s. and average values.

[84.8 V, 76.4 V]

9 A sinusoidal current has a mean value of15.0 A. Determine its maximum and r.m.s.values. [23.55 A, 16.65 A]

14.5 The equation of a sinusoidalwaveform

In Fig. 14.8, 0A represents a vector that is free torotate anticlockwise about 0 at an angular velocityof ω rad/s. A rotating vector is known as a phasor.

After time t seconds the vector 0A has turnedthrough an angle ωt. If the line BC is constructedperpendicular to 0A as shown, then

sin ωt DBC

0Bi.e. BC D 0B sin ωt

Figure 14.8

If all such vertical components are projected on toa graph of y against angle ωt (in radians), a sinecurve results of maximum value 0A. Any quantitywhich varies sinusoidally can thus be represented asa phasor.

A sine curve may not always start at 0°. Toshow this a periodic function is represented byy D sin⊲ωtš�⊳, where � is the phase (or angle) dif-ference compared with y D sin ωt. In Fig. 14.9(a),y2 D sin⊲ωt C �⊳ starts � radians earlier thany1 D sin ωt and is thus said to lead y1 by � radians.Phasors y1 and y2 are shown in Fig. 14.9(b) at thetime when t D 0.

Figure 14.9

In Fig. 14.9(c), y4 D sin⊲ωt �⊳ starts � radianslater than y3 D sin ωt and is thus said to lag y3 by �radians. Phasors y3 and y4 are shown in Fig. 14.9(d)at the time when t D 0.

Given the general sinusoidal voltage,v= V m sin.wt ± f/, then

(i) Amplitude or maximum value D Vm

(ii) Peak to peak value D 2 Vm

(iii) Angular velocity D ω rad/s(iv) Periodic time, T D 2�/ω seconds(v) Frequency, f D ω/2� Hz (since ω D 2�f)

(vi) � D angle of lag or lead (compared withv D Vm sin ωt)

Problem 9. An alternating voltage is givenby v D 282.8 sin 314 t volts. Find (a) ther.m.s. voltage, (b) the frequency and (c) theinstantaneous value of voltage whent D 4 ms.

(a) The general expression for an alternating voltageis v D Vm sin⊲ωt š �⊳. Comparing

TLFeBOOK


v D 282.8 sin 314 t with this general expressiongives the peak voltage as 282.8 V. Hence ther.m.s. voltage D 0.707ðmaximum valueD 0.707ð 282.8 D 200 V

(b) Angular velocity, ω D 314 rad/s, i.e. 2�f D

314. Hence frequency,

f D314

2�D 50 Hz

(c) When t D 4 ms,

v D 282.8 sin⊲314ð 4ð 10 3⊳

D 282.8 sin⊲1.256⊳ D 268.9 V

Note that 1.256 radians D

[

1.256ð180°

�

]

D 71.96°

Hence v D 282.8 sin 71.96° D 268.9 V, asabove.

Problem 10. An alternating voltage is givenby v D 75 sin⊲200�t 0.25⊳ volts. Find(a) the amplitude, (b) the peak-to-peak value,(c) the r.m.s. value, (d) the periodic time,(e) the frequency, and (f) the phase angle (indegrees and minutes) relative to 75 sin 200�t.

Comparing v D 75 sin⊲200�t 0.25⊳ with the gen-eral expression v D Vm sin⊲ωt š �⊳ gives:

(a) Amplitude, or peak value D 75 V

(b) Peak-to-peak value D 2ð 75 D 150 V

(c) The r.m.s. value D 0.707ðmaximum value

D 0.707ð 75 D 53 V

(d) Angular velocity, ω D 200� rad/s. Hence peri-odic time,

T D2�

ωD

2�

200�D

1

100D 0.01 s or 10 ms

(e) Frequency, f D1

TD

1

0.01D 100 Hz

(f) Phase angle, �D0.25 radians lagging

75 sin 200�t

0.25 rads D 0.25ð180°

�D 14.32°

Hence phase angle D 14.32° lagging

Problem 11. An alternating voltage, v, hasa periodic time of 0.01 s and a peak value of40 V. When time t is zero, v D 20 V.Express the instantaneous voltage in the formv D Vm sin⊲ωt š �⊳.

Amplitude, Vm D 40 V.

Periodic time T D2�

ωhence angular velocity,

ω D2�

TD

2�

0.01D 200� rad/s.

v D Vm sin⊲ωt C �⊳ thus becomes

v D 40 sin⊲200�t C �⊳ volts.

When time t D 0, v D 20 Vi.e. 20 D 40 sin �so that sin �D 20/40 D 0.5

Hence � D sin 1⊲ 0.5⊳ D 30°

D

(

30ð�

180

)

rads D �

6rads

Thus v= 40 sin

(

200pt −

p

6

)

V

Problem 12. The current in an a.c. circuit atany time t seconds is given by:i D 120 sin⊲100�t C 0.36⊳ amperes. Find:(a) the peak value, the periodic time, thefrequency and phase angle relative to120 sin 100�t (b) the value of the currentwhen t D 0 (c) the value of the current whent D 8 ms (d) the time when the current firstreaches 60 A, and (e) the time when thecurrent is first a maximum.

(a) Peak value D 120 A

Periodic time T D2�

ω

D2�

100�⊲since ω D 100�⊳

D1

50D 0.02 s or 20 ms

Frequency, f D1

TD

1

0.02D 50 Hz

TLFeBOOK


Phase angle D 0.36 rads

D 0.36ð180°

�D 20.63° leading

(b) When t D 0,

i D 120 sin⊲0C 0.36⊳

D 120 sin 20.63° D 42.3 A

(c) When t D 8 ms,

i D 120 sin

[

100�

(

8

103

)

C 0.36

]

D 120 sin 2.8733⊲D 120 sin 164.63°⊳

D 31.8 A

(d) When i D 60 A, 60 D 120 sin⊲100�t C 0.36⊳thus ⊲60/120⊳ D sin⊲100�t C 0.36⊳ so that

⊲100�t C 0.36⊳ D sin 1 0.5 D 30°

D �/6 rads D 0.5236 rads. Hence time,

t D0.5236 0.36

100�D 0.521 ms

(e) When the current is a maximum, i D 120 A.

Thus 120 D 120 sin⊲100�t C 0.36⊳

1 D sin⊲100�t C 0.36⊳

⊲100�t C 0.36⊳ D sin 1 1 D 90°

D ⊲�/2⊳ rads

D 1.5708 rads.

Hence time, t D1.5708 0.36

100�D 3.85 ms


Exercise 75 Further problems onv = Vm sin.wt ± f/

1 An alternating voltage is represented by v D

20 sin 157.1 t volts. Find (a) the maximumvalue (b) the frequency (c) the periodic time.(d) What is the angular velocity of the phasorrepresenting this waveform?

[(a) 20 V (b) 25 Hz(c) 0.04 s (d) 157.1 rads/s]

2 Find the peak value, the r.m.s. value, the peri-odic time, the frequency and the phase angle

(in degrees) of the following alternating quan-tities:(a) v D 90 sin 400�t volts

[90 V, 63.63 V, 5 ms, 200 Hz, 0°](b) i D 50 sin⊲100�t C 0.30⊳ amperes

[50 A, 35.35 A, 0.02 s, 50 Hz, 17.19° lead](c) e D 200 sin⊲628.4 t 0.41⊳ volts

[200 V, 141.4 V, 0.01 s, 100 Hz, 23.49°

lag]

3 A sinusoidal current has a peak value of 30 Aand a frequency of 60 Hz. At time t D 0,the current is zero. Express the instantaneouscurrent i in the form i D Im sin ωt

[i D 30 sin 120�t]

4 An alternating voltage v has a periodic timeof 20 ms and a maximum value of 200 V.When time t D 0, v D 75 volts. Deducea sinusoidal expression for v and sketch onecycle of the voltage showing important points.

[v D 200 sin⊲100�t 0.384⊳]

5 The voltage in an alternating current circuit atany time t seconds is given by v D 60 sin 40tvolts. Find the first time when the voltage is(a) 20 V (b) 30 V

[(a) 8.496 ms (b) 91.63 ms]

6 The instantaneous value of voltage in an a.c.circuit at any time t seconds is given byv D 100 sin⊲50�t 0.523⊳ V. Find:(a) the peak-to-peak voltage, the periodic

time, the frequency and the phase angle(b) the voltage when t D 0(c) the voltage when t D 8 ms(d) the times in the first cycle when the voltage

is 60 V(e) the times in the first cycle when the voltage

is 40 V(f) the first time when the voltage is a maxi-

mum.Sketch the curve for one cycle showingrelevant points. [(a) 200 V, 0.04 s, 25 Hz,29.97° lagging (b) 49.95 V (c) 66.96 V(d) 7.426 ms, 19.23 ms (e) 25.95 ms, 40.71 ms(f) 13.33 ms]

14.6 Combination of waveforms

The resultant of the addition (or subtraction) of twosinusoidal quantities may be determined either:

TLFeBOOK


(a) by plotting the periodic functions graphically(see worked Problems 13 and 16), or

(b) by resolution of phasors by drawing or calcula-tion (see worked Problems 14 and 15)

Problem 13. The instantaneous values oftwo alternating currents are given byi1 D 20 sin ωt amperes andi2 D 10 sin⊲ωt C �/3⊳ amperes. By plottingi1 and i2 on the same axes, using the samescale, over one cycle, and adding ordinates atintervals, obtain a sinusoidal expression fori1 C i2.

i1 D 20 sin ωt and i2 D 10 sin⊲ωt C �/3⊳ are shownplotted in Fig. 14.10. Ordinates of i1 and i2 areadded at, say, 15° intervals (a pair of dividers areuseful for this). For example,

at 30°, i1 C i2 D 10C 10 D 20 Aat 60°, i1 C i2 D 17.3C 8.7 D 26 Aat 150°, i1 C i2 D 10C ⊲ 5⊳ D 5 A, and so on.

Figure 14.10

The resultant waveform for i1 C i2 is shown by thebroken line in Fig. 14.10. It has the same period,and hence frequency, as i1 and i2. The amplitude orpeak value is 26.5 A

The resultant waveform leads the curve i1 D

20 sin ωt by 19° i.e. ⊲19ð �/180⊳ rads D 0.332 radsHence the sinusoidal expression for the resultant

i1 C i2 is given by:

iR = i1 + i2 = 26.5 sin.wt + 0.332/ A

Problem 14. Two alternating voltages arerepresented by v1 D 50 sin ωt volts andv2 D 100 sin⊲ωt �/6⊳ V. Draw the phasordiagram and find, by calculation, a sinusoidalexpression to represent v1 C v2.

Phasors are usually drawn at the instant when timet D 0. Thus v1 is drawn horizontally 50 unitslong and v2 is drawn 100 units long lagging v1 by�/6 rads, i.e. 30°. This is shown in Fig. 14.11(a)where 0 is the point of rotation of the phasors.

Figure 14.11

Procedure to draw phasor diagram to representv1 C v2:

(i) Draw v1 horizontal 50 units long, i.e. oa ofFig. 14.11(b)

(ii) Join v2 to the end of v1 at the appropriate angle,i.e. ab of Fig. 14.11(b)

(iii) The resultant vR D v1 C v2 is given by thelength ob and its phase angle may be measuredwith respect to v1

Alternatively, when two phasors are being added theresultant is always the diagonal of the parallelogram,as shown in Fig. 14.11(c).

From the drawing, by measurement, vR D 145 Vand angle � D 20° lagging v1.

TLFeBOOK


A more accurate solution is obtained by calcu-lation, using the cosine and sine rules. Using thecosine rule on triangle 0ab of Fig. 14.11(b) gives:

v2R D v

21 C v

22 2v1v2 cos 150°

D 502 C 1002 2⊲50⊳⊲100⊳ cos 150°

D 2500C 10000 ⊲ 8660⊳

vR Dp

21160 D 145.5 V

Using the sine rule,

100

sin �D

145.5

sin 150°

from which sin � D100 sin 150°

145.5

D 0.3436

and � D sin 1 0.3436 D 20.096° D 0.35 radians,and lags v1. Hence

vR D v1 C v2 D 145.5 sin.wt − 0.35/ V

Problem 15. Find a sinusoidal expressionfor ⊲i1 C i2⊳ of Problem 13, (b) by drawingphasors, (b) by calculation.

(a) The relative positions of i1 and i2 at time t D 0are shown as phasors in Fig. 14.12(a). The pha-sor diagram in Fig. 14.12(b) shows the resultantiR, and iR is measured as 26 A and angle � as19° or 0.33 rads leading i1.

Hence, by drawing, iR = 26 sin.wt + 0.33/ A

Figure 14.12

(b) From Fig. 14.12(b), by the cosine rule:

i2R D 202 C 102 2⊲20⊳⊲10⊳⊲cos 120°⊳

from which iR D 26.46 A

By the sine rule:

10

sin �D

26.46

sin 120°

from which � D 19.10° ⊲i.e. 0.333 rads⊳

Hence, by calculation,

iR = 26.46 sin.wt + 0.333/ A

Problem 16. Two alternating voltages aregiven by v1 D 120 sin ωt volts andv2 D 200 sin⊲ωt �/4⊳ volts. Obtainsinusoidal expressions for v1 v2 (a) byplotting waveforms, and (b) by resolution ofphasors.

(a) v1 D 120 sin ωt and v2 D 200 sin⊲ωt �/4⊳ areshown plotted in Fig. 14.13 Care must be takenwhen subtracting values of ordinates especiallywhen at least one of the ordinates is negative.For example

at 30°, v1 v2 D 60 ⊲ 52⊳ D 112 Vat 60°, v1 v2 D 104 52 D 52 Vat 150°, v1 v2 D 60 193 D 133 V andso on.

Figure 14.13

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The resultant waveform, vR D v1 v2, is shownby the broken line in Fig. 14.13 The maximumvalue of vR is 143 V and the waveform is seento lead v1 by 99° (i.e. 1.73 radians)

Hence, by drawing,

vR = v1 − v2 = 143 sin.wt + 1.73/volts

(b) The relative positions of v1 and v2 are shown attime t D 0 as phasors in Fig. 14.14(a). Sincethe resultant of v1 v2 is required, v2 isdrawn in the opposite direction to Cv2 and isshown by the broken line in Fig. 14.14(a). Thephasor diagram with the resultant is shown inFig. 14.14(b) where v2 is added phasoriallyto v1.

Figure 14.14

By resolution:

Sum of horizontal components of v1 and v2 D

120 cos 0° 200 cos 45° D 21.42

Sum of vertical components of v1 and v2 D

120 sin 0° C 200 sin 45° D 141.4

From Fig. 14.14(c), resultant

vR D

√

⊲ 21.42⊳2 C ⊲141.4⊳2

D 143.0

and tan �0 D141.4

21.42

D tan 6.6013

from which, �0 D tan 1 6.6013

D 81.39°

and � D 98.61° or 1.721 radians

Hence, by resolution of phasors,

vR = v1 − v2 = 143.0 sin.wt + 1.721/ volts


Exercise 76 Further problems on thecombination of periodic functions

1 The instantaneous values of two alternatingvoltages are given by v1 D 5 sin ωt and v2 D

8 sin⊲ωt �/6⊳. By plotting v1 and v2 on thesame axes, using the same scale, over onecycle, obtain expressions for(a) v1 C v2 and (b) v1 v2

[(a) v1 C v2 D 12.58 sin⊲ωt 0.325⊳ V(b) v1 v2 D 4.44 sin⊲ωt C 2.02⊳ V]

2 Repeat Problem 1 using resolution of phasors

3 Construct a phasor diagram to represent i1C i2

where i1 D 12 sin ωt andi2 D 15 sin⊲ωt C �/3⊳. By measurement, orby calculation, find a sinusoidal expression torepresent i1 C i2

[23.43 sin⊲ωt C 0.588⊳]

Determine, either by plotting graphs andadding ordinates at intervals, or by calculation,the following periodic functions in the formv D Vm sin⊲ωt š �⊳

4 10 sin ωt C 4 sin⊲ωt C �/4⊳[13.14 sin⊲ωt C 0.217⊳]

5 80 sin⊲ωt C �/3⊳C 50 sin⊲ωt �/6⊳[94.34 sin⊲ωt C 0.489⊳]

6 100 sin ωt 70 sin⊲ωt �/3⊳[88.88 sin⊲ωt C 0.751⊳]

14.7 Rectification

The process of obtaining unidirectional currents andvoltages from alternating currents and voltages iscalled rectification. Automatic switching in circuitsis carried out by devices called diodes. Half and full-wave rectifiers are explained in Chapter 11, Sec-tion 11.7, page 132

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Now try the following exercises

Exercise 77 Short answer questions onalternating voltages and currents

1 Briefly explain the principle of operation ofthe simple alternator

2 What is meant by (a) waveform (b) cycle

3 What is the difference between an alternatingand a unidirectional waveform?

4 The time to complete one cycle of a wave-form is called the . . . . . .

5 What is frequency? Name its unit

6 The mains supply voltage has a special shapeof waveform called a . . . . . .

7 Define peak value

8 What is meant by the r.m.s. value?

9 The domestic mains electricity voltage inGreat Britain is . . . . . .

10 What is the mean value of a sinusoidal alter-nating e.m.f. which has a maximum value of100 V?

11 The effective value of a sinusoidal waveformis . . . . . . ð maximum value

12 What is a phasor quantity?

13 Complete the statement:Form factor D . . . . . . ł . . . . . ., and for a sinewave, form factor D . . . . . .

14 Complete the statement:Peak factor D . . . . . . ł . . . . . ., and for a sinewave, peak factor D . . . . . .

15 A sinusoidal current is given by i D

Im sin⊲ωt š ˛⊳. What do the symbols Im, ω

and ˛ represent?

16 How is switching obtained when convertinga.c. to d.c.?

Exercise 78 Multi-choice questions onalternating voltages and currents (Answerson page 375)

1 The value of an alternating current at anygiven instant is:

(a) a maximum value(b) a peak value(c) an instantaneous value(d) an r.m.s. value

2 An alternating current completes 100 cyclesin 0.1 s. Its frequency is:(a) 20 Hz (b) 100 Hz

(c) 0.002 Hz (d) 1 kHz

3 In Fig. 14.15, at the instant shown, the gen-erated e.m.f. will be:(a) zero(b) an r.m.s. value(c) an average value(d) a maximum value

Figure 14.15

4 The supply of electrical energy for a con-sumer is usually by a.c. because:(a) transmission and distribution are more

easily effected(b) it is most suitable for variable speed

motors(c) the volt drop in cables is minimal(d) cable power losses are negligible

5 Which of the following statements is false?(a) It is cheaper to use a.c. than d.c.(b) Distribution of a.c. is more convenient

than with d.c. since voltages may bereadily altered using transformers

(c) An alternator is an a.c. generator(d) A rectifier changes d.c. to a.c.

6 An alternating voltage of maximum value100 V is applied to a lamp. Which of thefollowing direct voltages, if applied to thelamp, would cause the lamp to light with thesame brilliance?

(a) 100 V (b) 63.7 V

(c) 70.7 V (d) 141.4 V

7 The value normally stated when referring toalternating currents and voltages is the:

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(a) instantaneous value(b) r.m.s. value(c) average value(d) peak value

8 State which of the following is false. For asine wave:(a) the peak factor is 1.414(b) the r.m.s. value is 0.707ð peak value(c) the average value is 0.637ð r.m.s. value(d) the form factor is 1.11

9 An a.c. supply is 70.7 V, 50 Hz. Which of thefollowing statements is false?(a) The periodic time is 20 ms(b) The peak value of the voltage is 70.7 V(c) The r.m.s. value of the voltage is 70.7 V(d) The peak value of the voltage is 100 V

10 An alternating voltage is given by v D

100 sin⊲50�t 0.30⊳ V.Which of the following statements is true?(a) The r.m.s. voltage is 100 V(b) The periodic time is 20 ms(c) The frequency is 25 Hz(d) The voltage is leading v D 100 sin 50�t

by 0.30 radians

11 The number of complete cycles of an alter-nating current occurring in one second isknown as:(a) the maximum value of the alternating

current(b) the frequency of the alternating current(c) the peak value of the alternating current(d) the r.m.s. or effective value

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alternating waveforms

waveforms c

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position c maximum flux

cycle of alternating

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time axis

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